If the roots of ` ax^(3) + bx^2 + cx + d=0` are in G.P then the roots of ` dx^3 - cx^2 + bx -a=0` are in

Options

A. A.P
B. G.P
C. H.P
D. A.G.P

Answer

To solve the problem, we will follow these steps:

### Step 1: Understand the Given Information
We are given a cubic polynomial \( ax^3 + bx^2 + cx + d = 0 \) whose roots \( \alpha, \beta, \gamma \) are in geometric progression (G.P).

### Step 2: Express the Roots in G.P
If the roots are in G.P, we can express them as:
- \( \alpha = a/r \)
- \( \beta = a \)
- \( \gamma = ar \)

for some \( a \) and \( r \).

### Step 3: Transform the Polynomial
We will transform the polynomial by substituting \( x \) with \( -\frac{1}{x} \):
\[ f\left(-\frac{1}{x}\right) = a\left(-\frac{1}{x}\right)^3 + b\left(-\frac{1}{x}\right)^2 + c\left(-\frac{1}{x}\right) + d \]
This simplifies to:
\[ f\left(-\frac{1}{x}\right) = -\frac{a}{x^3} + \frac{b}{x^2} - \frac{c}{x} + d \]

### Step 4: Clear the Denominator
To eliminate the fraction, multiply the entire equation by \( -x^3 \):
\[ -a + b x - c x^2 + d x^3 = 0 \]
Rearranging gives us:
\[ d x^3 - c x^2 + b x - a = 0 \]

### Step 5: Analyze the Roots of the New Polynomial
The roots of the new polynomial \( d x^3 - c x^2 + b x - a = 0 \) are \( -\frac{1}{\alpha}, -\frac{1}{\beta}, -\frac{1}{\gamma} \).

### Step 6: Determine the Nature of the New Roots
Since \( \alpha, \beta, \gamma \) are in G.P, their reciprocals \( \frac{1}{\alpha}, \frac{1}{\beta}, \frac{1}{\gamma} \) will also be in G.P. Therefore, the negative of these values, \( -\frac{1}{\alpha}, -\frac{1}{\beta}, -\frac{1}{\gamma} \), will also be in G.P.

### Conclusion
Thus, the roots of the polynomial \( d x^3 - c x^2 + b x - a = 0 \) are also in G.P.

### Final Answer
The roots of \( d x^3 - c x^2 + b x - a = 0 \) are in G.P.

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